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Forest Diagramsfor
Thompson’s Group
Jim Belk
Associative LawsConsider the following piecewise-linear
homeomorphism of :
Associative LawsThis homeomorphism is called the basic
associative law.
It corresponds to the operation
Associative LawsHere’s a different associative law.
It corresponds to .
Associative Laws
General Definition:
A dyadic subdivision of is obtained by repeatedly cutting intervals in half:
Associative LawsAn associative law is a PL-homeomorphism that
maps linearly between the intervals of two dyadic subdivisions.
Associative Laws
Associative LawsThompson’s Group is the group of all associative
laws.
Associative LawsThompson’s Group is the group of all associative
laws.
If , then:
• Every slope of is a power of 2.
• Every breakpoint of has dyadic rational coordinates.
The converse also holds.
2
½
1
(¼,½)
(½,¾)
Properties of
is an infinite, torsion-free group.
Properties of
is an infinite, torsion-free group.
is finitely generated:
Properties of
is an infinite, torsion-free group.
is finitely generated.
is finitely presented:
2 relations
Properties of
is an infinite, torsion-free group.
is finitely generated.
is finitely presented.
admits a complex with exactly two cells in each dimension.
Properties of is simple. Every proper quotient of
is abelian.
has exponential growth.
contains .
does not contain .
• Is amenable?
Tree Diagrams
Tree DiagramsWe can represent an associative law by a
pair of binary trees:
This is called a tree diagram.
Tree DiagramsUnfortunately, the tree diagram for an
element of is not unique.
Tree DiagramsUnfortunately, the tree diagram for an
element of is not unique.
Tree DiagramsUnfortunately, the tree diagram for an
element of is not unique.
Tree DiagramsUnfortunately, the tree diagram for an
element of is not unique.
We can always cancel opposing pairs of carets.
This is called a reduction of the tree diagram.
Multiplying Tree Diagrams
Multiplying Tree Diagrams
Multiplying Tree Diagrams
Multiplying Tree Diagrams
GeneratorsHere are the tree diagrams for the
generators:
The Action on
The Action on If we conjugate by the homeomorphism:
we get an action of on .
A dyadic subdivision of is obtained by repeatedly cutting intervals in half:
is the group of PL-homeomorphisms of that map linearly between the intervals of two dyadic subdivisions
The Action on
The Action on
What’s the point?
The generators become simpler.
Here’s the new picture for :
The Action on
What’s the point?
The generators become simpler.
And here’s :
Forest Diagrams
Forest DiagramsWe can represent an element of using a
pair of binary forests:
Forest Diagrams
This is called a forest diagram.
Forest Diagrams
This is called a forest diagram.
Here are the forest diagrams for the generators:
The Action of
Left-multiplication by moves the top pointer of a forest diagram:
The Action of
Left-multiplication by moves the top pointer of a forest diagram:
The Action of
Left-multiplication by moves the top pointer of a forest diagram:
The Action of
Left-multiplication by moves the top pointer of a forest diagram:
The Action of
Left-multiplication by moves the top pointer of a forest diagram:
The Action of
Left-multiplication by moves the top pointer of a forest diagram:
Here are the forest diagrams for the generators:
The Action of
Left-multiplication by adds a new caret on the top:
The Action of
Left-multiplication by adds a new caret on the top:
The Action of
Left-multiplication by adds a new caret on the top:
The Action of
Left-multiplication by adds a new caret on the top:
The Action of
Sometimes the new caret appears opposite a caret on the bottom:
The Action of
Sometimes the new caret appears opposite a caret on the bottom:
The Action of
Sometimes the new caret appears opposite a caret on the bottom:
The Action of
So can delete bottom carets (and can
create bottom carets.)
Lengths
Finding Lengths
Problem. Given an , find the -length of .
Example. Find the length of:
Finding Lengths
Finding Lengths
1
Finding Lengths
1
Finding Lengths
2
Finding Lengths
21
Finding Lengths
211
Finding Lengths
211
Finding Lengths
2111
Finding Lengths
2121
Finding Lengths
2221
Finding Lengths
2221
(1 + 2 + 2 + 2) + 2 = 9
This element has length 9.
Example 2
Example. Find the length of:
Example 2
Example 2
1
Example 2
1 1
Example 2
1 1
Example 2
1 1 1
Example 2
1 1 1 1
Example 2
1 1 1 1
Example 2
1 1 1 2
Example 2
1 1 2 2
Example 2
1 2 2 2
Example 2
2 2 2 2
Example 2
2 2 2 2
(2 + 2 + 2 + 2) + 2 = 10
This element has length 10.
Example 3
Example. Find the length of:
Example 3
Example 3
1
Example 3
1 1
Example 3
1 1
Example 3
1 1 1
Example 3
1 1 1
Example 3
1 1 2
Example 3
1 1 2
Example 3
1 2 2
Example 3
1 2 2
Example 3
2 2 2
Example 3
2 2 21
Example 3
2 2 21
(1 + 2 + 2 + 2) + 4 = 11
This element has length 11.
Labeling. Label each space as follows.
The Length Formula
Labeling. Label each space as follows.
L: Left of the pointer, exterior.
The Length Formula
L
Labeling. Label each space as follows.
L: Left of the pointer, exterior.
N: Next to a caret on the left.
The Length Formula
L NN N
Labeling. Label each space as follows.
L: Left of the pointer, exterior.
N: Next to a caret on the left.
R: Right of the pointer, exterior.
The Length Formula
L N NNR
Labeling. Label each space as follows.
L: Left of the pointer, exterior.
N: Next to a caret on the left.
R: Right of the pointer, exterior.
I: Interior
The Length Formula
L N R NNI I I
The Length Formula
Theorem. The weight of a space is determined by its label pair.
L N R I
L 2 1 1 1
N 1 2 2 2
R 1 2 2 0
I 1 2 0 0
L: Left exterior.
N: Next to a caret.
R: Right exterior.
I: Interior
L N R I
L 2 1 1 1
N 1 2 2 2
R 1 2 2 0
I 1 2 0 0
L: Left exterior.
N: Next to a caret.
R: Right exterior.
I: Interior
L N R I
L 2 1 1 1
N 1 2 2 2
R 1 2 2 0
I 1 2 0 0
L: Left exterior.
N: Next to a caret.
R: Right exterior.
I: Interior
L
L N R I
L 2 1 1 1
N 1 2 2 2
R 1 2 2 0
I 1 2 0 0
L: Left exterior.
N: Next to a caret.
R: Right exterior.
I: Interior
NL N
N
L N R I
L 2 1 1 1
N 1 2 2 2
R 1 2 2 0
I 1 2 0 0
L: Left exterior.
N: Next to a caret.
R: Right exterior.
I: Interior
LR
NN NR RR R R R
L N R I
L 2 1 1 1
N 1 2 2 2
R 1 2 2 0
I 1 2 0 0
L: Left exterior.
N: Next to a caret.
R: Right exterior.
I: Interior
L NN N II
IR R R
R R R R
L N R I
L 2 1 1 1
N 1 2 2 2
R 1 2 2 0
I 1 2 0 0
L: Left exterior.
N: Next to a caret.
R: Right exterior.
I: Interior
L NN N
1
R R RR R R RI
I I
2 0 2 0 2 0
L NN NR R RR R R RI
I I
1 2 0 2 0 2 0
length weights + # of carets 7 + 4
11
Convexity
ConvexityA group is convex if is convex for each
.
id
Convexity is almost convex if any two elements in
a distance two apart are connected by a path of length .
id
ConvexityTheorem (Cleary and Taback). is not
almost convex (using ).
ConvexityTheorem (Cleary and Taback). is not
almost convex (using ).
1 1 2 2 2 2 2 2 0
Length 16
ConvexityTheorem (Cleary and Taback). is not
almost convex (using ).
1 2 2 2 2 2 2 2 0
Length 17
ConvexityTheorem (Cleary and Taback). is not
almost convex (using ).
0 2 2 2 2 2 2 2 0
Length 16
ConvexityTheorem (Cleary and Taback). is not
almost convex (using ).
1 2 2 2 2 2 2 2 0
Length 17
ConvexityTheorem (Cleary and Taback). is not
almost convex (using ).
1 1 2 2 2 2 2 2 0
Length 16
ConvexityTheorem (Cleary and Taback). is not
almost convex (using ).
ConvexityTheorem (Cleary and Taback). is not
almost convex (using ).
This gives us two elements of a distance two apart that have distance in .
Convexity
Theorem (Belk and Bux). For even, there exist elements such that:
, and
• The shortest path in from to has length .
Kai-Uwe and I have proven the following:
Amenability
The Isoperimetric Constant
Let be the Cayley graph of a group .
If is a finite subset of ,
its boundary consists of
all edges between and
.
The Isoperimetric Constant
Let be the Cayley graph of a group .
The isoperimetric constant is:
is amenable if .
Theorem (Belk and Brown). .Proof: Let be all elements of the form:
We claim that:
Given a random , we must compute:
exits the current tree of is trivial.
Claim. Given a random element of :
current tree is trivial
as .
Theorem. As , the probability that the current
tree is trivial satisfies:
where is the number of binary trees with
leaves.
Theorem. As , the probability that the current
tree is trivial satisfies:
where is the number of binary trees with
leaves.
The coefficients are the Catalan numbers, and have growth rate .
The polynomial above “converges” to the generating function for the Catalan numbers, which has a vertical asymptote at .
The End